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Math-564-A1 · revision r1 · on Math-564

gpt 6 astra ultra attempt on Math-564

Exploration · LLM-generated · gpt 6 astra ultra ·

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Contents
  1. Infinitely Many Leinster Groups
  2. Definitions and mathematical statement
  3. Shared notation.
  4. Statement recorded in the supplied source.
  5. Residual task reported by the archive.
  6. Short English statement
  7. Research attempt: coprime products and arithmetic obstructions
  8. 1. Cyclic groups and a conditional infinite supply
  9. 2. Normal subgroup sums multiply under a precise hypothesis
  10. 3. Excluding tempting factors
  11. 4. Verification and what remains
  12. Sources

Infinitely Many Leinster Groups

Definitions and mathematical statement

Shared notation.
Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

For a finite group $G$, let $\sigma_N(G)=\sum_{H\trianglelefteq G}|H|$, including both the identity subgroup and $G$. A Leinster group satisfies $\sigma_N(G)=2|G|$. Infinitude means infinitely many isomorphism classes of finite groups with this property, not infinitely many differently labelled copies of one group. [S1] [S2]

Statement recorded in the supplied source.
Are there infinitely many Leinster groups? [S1]
Residual task reported by the archive.

The embedded review $2026-08-17$ records: Construct an infinite family or prove only finitely many exist. [S1]

Short English statement

Are there infinitely many finite groups for which the orders of all normal subgroups add up to twice the group’s order?

Research attempt: coprime products and arithmetic obstructions

Status. No infinite family is obtained. The inspected abstract of Leinster's original paper [S2] connects cyclic examples with perfect numbers. We prove that connection and a product criterion, then identify precisely why they do not yield unconditional infinitude.

1. Cyclic groups and a conditional infinite supply

A cyclic group $C_n$ has one subgroup of each divisor order $d\mid n$, all normal. Hence $\sigma_N(C_n)=\sigma(n)=\sum_{d\mid n}d$, and $C_n$ is Leinster exactly when $n$ is a perfect number. If $q=2^p-1$ is prime, then $p>1$, $q$ is odd, and coprime divisor factorization gives \[\sigma(2^{p-1}q)=(1+2+\cdots+2^{p-1})(1+q) =(2^p-1)2^p=2(2^{p-1}q).\] Thus infinitely many Mersenne primes would imply infinitely many pairwise nonisomorphic Leinster groups. The implication is proved; its antecedent is not proved here. The first orders from $p=2,3,5,7$ are $6,28,496,8128$, which the diagnostic verifies by direct divisor sums.

2. Normal subgroup sums multiply under a precise hypothesis

Let $G,H$ have coprime orders. Every subgroup $L\leq G\times H$ is a product. To see this, for $(g,h)\in L$ choose an integer $e$ congruent to 1 modulo $|G|$ and 0 modulo $|H|$. Then $(g,h)^e=(g,1)\in L$; a complementary exponent gives $(1,h)\in L$. Thus $L=L_G\times L_H$. Such a product is normal exactly when each factor is normal, as follows by conjugating separately in each coordinate. Consequently \[\sigma_N(G\times H)=\sigma_N(G)\sigma_N(H),\qquad \rho(G\times H)=\rho(G)\rho(H),\quad \rho(G)=\frac{\sigma_N(G)}{|G|}.\tag{1}\] This supplies an exact construction test: coprime factors must have ratios multiplying to 2. It does not allow repeated multiplication of existing examples; the product of two coprime Leinster groups has ratio 4.

3. Excluding tempting factors

No nontrivial finite $p$-group is Leinster. Its identity subgroup contributes 1 to $\sigma_N$, whereas every other subgroup order is divisible by $p$. Thus $\sigma_N\equiv1\pmod p$, while $2|G|\equiv0\pmod p$. For a nonabelian finite simple group $S$, only $1$ and $S$ are normal, so $\rho(S)=1+1/|S|<2$; it is not itself a solution either. Appending a coprime cyclic prime-order factor multiplies a ratio by $1+1/p>1$, and therefore cannot preserve ratio 2.

The coprimality hypothesis in $1$ is indispensable. For example, $C_p\times C_p$ has $p+1$ subgroups of order $p$, so \[\sigma_N(C_p\times C_p)=1+(p+1)p+p^2=1+p+2p^2, \] whereas $\sigma_N(C_p)^2=(1+p)^2$. Their difference is $p(p-1)>0$. The extra diagonal subgroups invalidate naive multiplicativity when orders overlap.

4. Verification and what remains

The diagnostic checks cyclic perfect orders below 10000, the displayed elementary-abelian calculation for $p=2,3,5,7$, and coprime cyclic multiplicativity for small orders. These tests confirm the arithmetic formulas rather than the existence of infinitely many groups. An unconditional infinite sequence of coprime ratio factorizations, or a construction controlling the additional normal subgroups when factors share primes, remains absent. Reducing infinitude to an unproved prime-producing assertion is not a solution of the original target.

Sources